Question Number 210112 by klipto last updated on 31/Jul/24
$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}} }{\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{b}}\right)}\boldsymbol{\mathrm{dx}}=\frac{\left(\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}}\right)^{\boldsymbol{\mathrm{n}}} −\mathrm{1}}{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{a}}}\boldsymbol{\pi\mathrm{csc}}\left(\boldsymbol{\pi\mathrm{n}}\right)\:\boldsymbol{\mathrm{a}}>\mathrm{0},\boldsymbol{\mathrm{b}}>\mathrm{0},\mid\boldsymbol{\mathrm{n}}\mid<\mathrm{1} \\ $$$$\boldsymbol{\mathrm{guys}}\:\boldsymbol{\mathrm{kill}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{me}}\:\boldsymbol{\mathrm{see}} \\ $$
Answered by Spillover last updated on 02/Aug/24
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{\left({x}+\mathrm{1}\right)\left({ax}+{b}\right)}{dx} \\ $$$${u}=\frac{{x}^{{n}} }{{ax}+{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{dv}=\frac{{dx}}{{x}+\mathrm{1}} \\ $$$$\int{udv}={uv}−\int{vdu}\:\: \\ $$$$\:\:\:\:\mid\frac{{x}^{{n}} \mathrm{ln}\:\left({x}+\mathrm{1}\right)}{{ax}+{b}}\mid_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \mathrm{ln}\:\left(\mathrm{1}+{x}\right)\frac{{d}}{{dx}}\left(\frac{{x}^{{n}} }{{ax}+{b}}\:\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{x}^{{n}} \mathrm{ln}\:\left({x}+\mathrm{1}\right)}{{ax}+{b}}\:\:\:\:\:\:\:\:\:{at}\:{x}=\mathrm{0} \\ $$$$\frac{{x}^{{n}} \mathrm{ln}\:\left({x}+\mathrm{1}\right)}{{ax}+{b}}=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{ln}\:\left(\mathrm{1}+{x}\right)\frac{{d}}{{dx}}\left(\frac{{x}^{{n}} }{{ax}+{b}}\:\right) \\ $$$${something}\:{wrong} \\ $$$$…… \\ $$
Answered by Spillover last updated on 02/Aug/24
![∫_0 ^∞ (x^n /((x+1)(ax+b)))dx (x^n /((x+1)(ax+b)))=(A/(x+1))+(B/(ax+b)) x^n =A(ax+b)+B(x+1) x^n =(Aa+B)x+B(Ab+B) (Aa+B)=0 (Ab+B)=x^n A=(b^n /(b−a)) B=−(a^n /(b−a)) ∫_0 ^∞ ((b^n /(b−a))/(x+1))dx−∫_0 ^∞ ((a^n /(b−a))/(ax+b))dx (b^n /(b−a)) ∫_0 ^∞ (1/(x+1))dx−(a^n /(b−a))∫_0 ^∞ (dx/(ax+b)) (b^n /(b−a)) ∫_0 ^∞ (1/(x+1))dx−(a^n /(b−a))∫_0 ^∞ (dx/(ax+b)) (b^n /(b−a)) πcosec (πn)−(a^n /(b−a))πcosec (πn) πcosec (πn)[(b^n /(b−a))−(a^n /(b−a))] πcosec (πn)(((b^n −a^n )/(b−a))) πcosec (πn)[((((b/a))^n −1)/(b−a))]](https://www.tinkutara.com/question/Q210204.png)
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{\left({x}+\mathrm{1}\right)\left({ax}+{b}\right)}{dx} \\ $$$$\frac{{x}^{{n}} }{\left({x}+\mathrm{1}\right)\left({ax}+{b}\right)}=\frac{{A}}{{x}+\mathrm{1}}+\frac{{B}}{{ax}+{b}} \\ $$$${x}^{{n}} ={A}\left({ax}+{b}\right)+{B}\left({x}+\mathrm{1}\right) \\ $$$${x}^{{n}} =\left({Aa}+{B}\right){x}+{B}\left({Ab}+{B}\right) \\ $$$$\left({Aa}+{B}\right)=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\left({Ab}+{B}\right)={x}^{{n}} \\ $$$${A}=\frac{{b}^{{n}} }{{b}−{a}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{B}=−\frac{{a}^{{n}} }{{b}−{a}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\frac{{b}^{{n}} }{{b}−{a}}}{{x}+\mathrm{1}}{dx}−\int_{\mathrm{0}} ^{\infty} \frac{\frac{{a}^{{n}} }{{b}−{a}}}{{ax}+{b}}{dx} \\ $$$$\frac{{b}^{{n}} }{{b}−{a}}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}+\mathrm{1}}{dx}−\frac{{a}^{{n}} }{{b}−{a}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{ax}+{b}} \\ $$$$\frac{{b}^{{n}} }{{b}−{a}}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}+\mathrm{1}}{dx}−\frac{{a}^{{n}} }{{b}−{a}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{ax}+{b}} \\ $$$$\frac{{b}^{{n}} }{{b}−{a}}\:\pi\mathrm{cosec}\:\left(\pi{n}\right)−\frac{{a}^{{n}} }{{b}−{a}}\pi\mathrm{cosec}\:\left(\pi{n}\right) \\ $$$$\pi\mathrm{cosec}\:\left(\pi{n}\right)\left[\frac{{b}^{{n}} }{{b}−{a}}−\frac{{a}^{{n}} }{{b}−{a}}\right] \\ $$$$\pi\mathrm{cosec}\:\left(\pi{n}\right)\left(\frac{{b}^{{n}} −{a}^{{n}} }{{b}−{a}}\right) \\ $$$$\pi\mathrm{cosec}\:\left(\pi{n}\right)\left[\frac{\left(\frac{{b}}{{a}}\right)^{{n}} −\mathrm{1}}{{b}−{a}}\right] \\ $$$$ \\ $$
Commented by klipto last updated on 02/Aug/24
$$\mathrm{spill}\:\mathrm{kilerbean}\:\mathrm{lol} \\ $$
Commented by klipto last updated on 02/Aug/24
$$\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{condition}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{n}} \\ $$